Mixing
Lineintegral $int_{gamma}|z|^2dz$ over ellipse
Let $a,binmathbb{R}_{>0}$ and $gamma: [0,2pi]rightarrowmathbb{C},tmapsto acos(t)+ibsin(t)$
calculate the line integral
$int_{gamma}|z|^2dz$
My calculation turns out to be really ugly. Is there maybe a "nice" way to calculate this integral?
integration complex-analysis complex-integration line-integrals
asked Nov 20 '18 at 15:49
Christian Singer
334213
add a comment |
Let $a,binmathbb{R}_{>0}$ and $gamma: [0,2pi]rightarrowmathbb{C},tmapsto acos(t)+ibsin(t)$
calculate the line integral
$int_{gamma}|z|^2dz$
My calculation turns out to be really ugly. Is there maybe a "nice" way to calculate this integral?
integration complex-analysis complex-integration line-integrals
asked Nov 20 '18 at 15:49
Christian Singer
334213
add a comment |
Let $a,binmathbb{R}_{>0}$ and $gamma: [0,2pi]rightarrowmathbb{C},tmapsto acos(t)+ibsin(t)$
calculate the line integral
$int_{gamma}|z|^2dz$
My calculation turns out to be really ugly. Is there maybe a "nice" way to calculate this integral?
integration complex-analysis complex-integration line-integrals
asked Nov 20 '18 at 15:49
Christian Singer
334213
Let $a,binmathbb{R}_{>0}$ and $gamma: [0,2pi]rightarrowmathbb{C},tmapsto acos(t)+ibsin(t)$
calculate the line integral
$int_{gamma}|z|^2dz$
My calculation turns out to be really ugly. Is there maybe a "nice" way to calculate this integral?
integration complex-analysis complex-integration line-integrals
integration complex-analysis complex-integration line-integrals
asked Nov 20 '18 at 15:49
Christian Singer
334213
asked Nov 20 '18 at 15:49
Christian Singer
334213
asked Nov 20 '18 at 15:49
Christian Singer
334213
asked Nov 20 '18 at 15:49
Christian Singer
334213
asked Nov 20 '18 at 15:49
Christian Singer
334213
334213
add a comment |
add a comment |
1 Answer
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The integrals are not ugly at all. You obtain
$$int_gamma|z|^2>dz=int_{omega-pi}^{omega+pi}bigl(acos^2 t+b^2sin^2 tbigr)(-asin t+ibcos t)>dt ,$$
whereby $omega$ can be chosen at will, due to periodicity. Choose $omega:=0$ for the real part, then $omega:={piover2}$ for the imaginary part, and note that the respective integrands are odd with respect to these points.
answered Nov 20 '18 at 16:22
Christian Blatter
172k7112326
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
The integrals are not ugly at all. You obtain
$$int_gamma|z|^2>dz=int_{omega-pi}^{omega+pi}bigl(acos^2 t+b^2sin^2 tbigr)(-asin t+ibcos t)>dt ,$$
whereby $omega$ can be chosen at will, due to periodicity. Choose $omega:=0$ for the real part, then $omega:={piover2}$ for the imaginary part, and note that the respective integrands are odd with respect to these points.
answered Nov 20 '18 at 16:22
Christian Blatter
172k7112326
add a comment |
The integrals are not ugly at all. You obtain
$$int_gamma|z|^2>dz=int_{omega-pi}^{omega+pi}bigl(acos^2 t+b^2sin^2 tbigr)(-asin t+ibcos t)>dt ,$$
whereby $omega$ can be chosen at will, due to periodicity. Choose $omega:=0$ for the real part, then $omega:={piover2}$ for the imaginary part, and note that the respective integrands are odd with respect to these points.
answered Nov 20 '18 at 16:22
Christian Blatter
172k7112326
add a comment |
The integrals are not ugly at all. You obtain
$$int_gamma|z|^2>dz=int_{omega-pi}^{omega+pi}bigl(acos^2 t+b^2sin^2 tbigr)(-asin t+ibcos t)>dt ,$$
whereby $omega$ can be chosen at will, due to periodicity. Choose $omega:=0$ for the real part, then $omega:={piover2}$ for the imaginary part, and note that the respective integrands are odd with respect to these points.
answered Nov 20 '18 at 16:22
Christian Blatter
172k7112326
The integrals are not ugly at all. You obtain
$$int_gamma|z|^2>dz=int_{omega-pi}^{omega+pi}bigl(acos^2 t+b^2sin^2 tbigr)(-asin t+ibcos t)>dt ,$$
whereby $omega$ can be chosen at will, due to periodicity. Choose $omega:=0$ for the real part, then $omega:={piover2}$ for the imaginary part, and note that the respective integrands are odd with respect to these points.
answered Nov 20 '18 at 16:22
Christian Blatter
172k7112326
answered Nov 20 '18 at 16:22
Christian Blatter
172k7112326
answered Nov 20 '18 at 16:22
Christian Blatter
172k7112326
answered Nov 20 '18 at 16:22
Christian Blatter
172k7112326
172k7112326
add a comment |
add a comment |
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